The total pressure exerted in ideal binary solution is given by `P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)` where `P_(A)^(@)&P_(B)^(@)` are the respective vapour pressure of pure components and `X_(A)&X_(B)` are their mole fraction in liquid phase. And composition of the vapour phase is determined with the help of Datton's law partial pressure: `Y_(A)=(P_(A)^(@)X_(A))/(P)`
If total pressure exerted in an ideal binary solution is given by `P=(5400)/(60+30Y_(A))mm` of Hg.
If the value of `Y_(A)=0.4` then the value of `X_(B)` is:

Options

A. `0.6`
B. `0.25`
C. `0.4`
D. `0.5`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given the equation for total pressure in an ideal binary solution:
\[ P = P_A^{(@)} X_A + P_B^{(@)} X_B \]
where \( P_A^{(@)} \) and \( P_B^{(@)} \) are the vapor pressures of pure components A and B, and \( X_A \) and \( X_B \) are their mole fractions in the liquid phase.

We also have the equation for the composition of the vapor phase:
\[ Y_A = \frac{P_A^{(@)} X_A}{P} \]

The total pressure is given by:
\[ P = \frac{5400}{60 + 30Y_A} \, \text{mm of Hg} \]

### Step 2: Calculate Total Pressure (P) when \( Y_A = 0.4 \)
Substituting \( Y_A = 0.4 \) into the equation for total pressure:
\[ P = \frac{5400}{60 + 30 \times 0.4} \]
Calculating the denominator:
\[ 60 + 30 \times 0.4 = 60 + 12 = 72 \]
Now, substituting back:
\[ P = \frac{5400}{72} = 75 \, \text{mm of Hg} \]

### Step 3: Determine Vapor Pressures \( P_A^{(@)} \) and \( P_B^{(@)} \)
From the problem, we know:
- When \( Y_A = 1 \) (pure A), \( P_A^{(@)} = 90 \, \text{mm of Hg} \)
- When \( Y_A = 0 \) (pure B), \( P_B^{(@)} = 5400 / 60 = 90 \, \text{mm of Hg} \)

### Step 4: Use the Total Pressure Equation
We know:
\[ P = P_A^{(@)} X_A + P_B^{(@)} X_B \]
Substituting \( P_A^{(@)} = 90 \) and \( P_B^{(@)} = 90 \):
\[ 75 = 90 X_A + 90 X_B \]
Since \( X_A + X_B = 1 \), we can express \( X_A \) as \( 1 - X_B \):
\[ 75 = 90(1 - X_B) + 90 X_B \]
This simplifies to:
\[ 75 = 90 - 90 X_B + 90 X_B \]
Thus, we can rearrange:
\[ 75 = 90 - 90 X_B \]
\[ 90 X_B = 90 - 75 \]
\[ 90 X_B = 15 \]
\[ X_B = \frac{15}{90} = \frac{1}{6} \]

### Step 5: Calculate \( X_B \)
Since \( X_B = \frac{1}{6} \), we can also find \( X_A \):
\[ X_A = 1 - X_B = 1 - \frac{1}{6} = \frac{5}{6} \]

### Final Answer
Thus, the value of \( X_B \) is:
\[ X_B = \frac{1}{6} \approx 0.167 \]